Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How does one compute (without brute force) the smallest integer $n$ such that

$\binom{2n}{1}(-3)^0 + \binom{2n}{3}(-3)^1 + \binom{2n}{5}(-3)^2 + \cdots + \binom{2n}{2n-1}(-3)^{(n-1)} = 0$?

share|improve this question
    
What does $C$ mean? –  Dimitrije Kostic Dec 1 '11 at 2:05
3  
You can use \binom{a}{b} if you mean the binomial coefficient. –  alex.jordan Dec 1 '11 at 2:10
1  
@jebyrnes - is this a hint or something else? –  Victor Dec 1 '11 at 2:19
3  
@Victor: I'm not trying to criticize your question. I'm just saying you'll get more helpful responses if you use notation that people recognize. Your question will just be easier to understand. –  Dimitrije Kostic Dec 1 '11 at 2:25
7  
I hope my substantial edit doesn't offend anyone. The question was unreadable as originally posted. –  user02138 Dec 1 '11 at 2:30

3 Answers 3

up vote 11 down vote accepted

$(1+i \sqrt{3})^{2n} = \binom{2n}{0} (i \sqrt{3})^0 + \binom{2n}{1} (i \sqrt{3})^1 + \binom{2n}{2} (i \sqrt{3})^2 + \cdots + \binom{2n}{2n} (i \sqrt{3})^{2n}$

$(1-i \sqrt{3})^{2n} = \binom{2n}{0} (i \sqrt{3})^0 + \binom{2n}{1} (-i \sqrt{3})^1 + \binom{2n}{2} (i \sqrt{3})^2 + \cdots + \binom{2n}{2n} (i \sqrt{3})^{2n}$

Subtract both to get,

$$\frac{(1+i \sqrt{3})^{2n} - (1-i \sqrt{3})^{2n}}{2} = \binom{2n}{1} (i \sqrt{3})^1 + \binom{2n}{3} (i \sqrt{3})^3 + \cdots + \binom{2n}{2n-1} (i \sqrt{3})^{2n-1}$$

$$\frac{(1+i \sqrt{3})^{2n} - (1-i \sqrt{3})^{2n}}{2 i \sqrt{3}} = \binom{2n}{1} + \binom{2n}{3} (i \sqrt{3})^2 + \binom{2n}{5} (i \sqrt{3})^4 + \cdots + \binom{2n}{2n-1} (i \sqrt{3})^{2n-2}$$

$$\frac{(1+i \sqrt{3})^{2n} - (1-i \sqrt{3})^{2n}}{2 i \sqrt{3}} = \frac{2^{2n}}{2i\sqrt{3}} \left(\cos \left(\frac{2n \pi}{3} \right) + i \sin \left(\frac{2n \pi}{3} \right) - \cos \left(\frac{2n \pi}{3} \right) + i \sin \left(\frac{2n \pi}{3} \right)\right)$$

Hence, we get $$\binom{2n}{1} + \binom{2n}{3} (- 3)^1 + \binom{2n}{5} (- 3)^2 + \cdots + \binom{2n}{2n-1} (- 3)^{n-1} = \frac{4^{n}}{\sqrt{3}} \sin \left( \frac{2n \pi}{3} \right)$$ Hence, whenever $n = \frac{3 k}{2}$ where $k \in \mathbb{Z}$, $$\binom{2n}{1} + \binom{2n}{3} (- 3)^1 + \binom{2n}{5} (- 3)^2 + \cdots + \binom{2n}{2n-1} (- 3)^{n-1} = 0 $$

share|improve this answer

Let's replace an explicit number $-3$ with a symbol $z$, so that we consider: $$ \sum_{k=1}^n z^{k-1} \binom{2n}{2k-1} = \sum_{m=0}^{2n} \left( \frac{1-(-1)^m}{2} \right)z^{(m-1)/2} \binom{2n}{m} = \frac{\left(1+\sqrt{z}\right)^{2n} - \left(1-\sqrt{z}\right)^{2n}}{2 \sqrt{z}} $$ Now substitute $z=-3$. Notice that $1 \pm \sqrt{-3} = 2 \exp\left( \pm i \frac{\pi}{3} \right)$, so it follows that the smallest such an integer is $n=3$.

share|improve this answer

Here is a hint. Prove the identity \begin{align} \sum_{i = 1}^{n} \binom{2n}{2i-1} x^{i-1} = \frac{(1 + 2 \sqrt{x} + x)^{n} - (1 - 2 \sqrt{x} + x)^{n} }{2 \sqrt{x}}. \end{align} Setting $x = -3$, we have \begin{align} \frac{(1 + 2 \sqrt{-3} -3)^{n} - (1 - 2 \sqrt{-3} -3)^{n} }{2 \sqrt{-3}} = \frac{4^{n}}{\sqrt{3}} \sin (\tfrac{2 \pi n}{3}) \end{align} Using an argument involving periodicity, you should conclude that this sum vanishes infinitely often and you can therefore determine the smallest $n$ accordingly, which is $n = 3$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.